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Change the order of integration and evaluate the following double integral:

[tex]I = {\int_0^{1} \left({\int\limits_{y}^{1}

30 y\sqrt{1+x^3} \mathrm{d}x }\right) {\mathrm{d}y} [/tex]

So thenn i did

[tex] = 30 \int_0^{1} \sqrt{1+x^3} \left({\int_0^{x} y \mathrm{d}y}\right) \mathrm{d}x [/tex]

[tex]= 30 \int_0^{1} \sqrt{1+x^3} \left(\frac{x^2}{2} \right) \mathrm{d}x \end{align}[/tex]

using integration by parts....

for [tex] \sqrt{1+x^3} [/tex]

[tex] let u = \sqrt{1+x^3} \qquad du= \frac{1}{2} \left(\sqrt{1+x^3}\right) 3x^2 = \frac{3x^2}{2\sqrt{1+x^3}} \qquad dv = dx \qquad v = x [/tex]

Thus!

[tex] = x \sqrt{1+x^3} - \int \frac{3x^3}{2\sqrt{1+x^3}} \mathrm{d}x [/tex]

after that.... i have no clue what to do. a lil help? thanks!

am i on the right track though?

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# Homework Help: Evaluate the following double integral

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